Welcome to today's mathematics lesson. We are going to explore Chapter 8: Logarithms. By the end of this session, you will understand what logarithms are, how to convert between exponential and logarithmic forms, the fundamental laws that govern logarithms, and some powerful techniques for simplifying complex expressions.
Let us begin with the fundamental idea. Logarithms were invented to make complicated calculations simpler. Think about the exponential statement: 3 raised to the power 4 equals 81. We write this as 3⁴ = 81. Now, here is the crucial insight: this same relationship between the three numbers 3, 4, and 81 can be expressed in a completely different way. We can ask: "To what power must 3 be raised to obtain 81?" The answer is 4. In logarithmic notation, we write: log₃ 81 = 4, which we read as "logarithm of 81 to the base 3 is 4."
Here is the formal definition you must remember. If a, b, and c are real numbers such that a is not equal to 1, and if aᵇ = c, then b is called the logarithm of c to the base a, written as logₐ c = b. The symbol ⇔ shows that these two forms are equivalent: aᵇ = c ⇔ logₐ c = b.
Let us practice converting between these two forms. Consider 2⁻³ = 0.125. This is the exponential form. Converting to logarithmic form: log₂ 0.125 = -3.
Conversely, if we are given log₆₄ 8 = ½, we convert to exponential form as 64^(½) = 8. Remember that a power of one half means square root.
Two special results emerge immediately from the definition. First, since any non-zero number raised to the power 0 equals 1, we have x⁰ = 1, which gives us logₓ 1 = 0. In general, the logarithm of 1 to any base is always zero.
Second, since any number raised to the power 1 equals itself, we have x¹ = x, which gives us logₓ x = 1. In general, the logarithm of any number to the same base is always one.
Let us work through some examples to build your confidence.
Example one: Find the logarithm of 1000 to the base 10. Let log₁₀ 1000 = x. By definition, this means 10ˣ = 1000. Since 1000 equals 10 cubed, we have 10ˣ = 10³. Therefore, x equals 3.
Example two: Find the logarithm of one-ninth to the base 3. Let log₃ (1/9) = x. This means 3ˣ = 1/9. Since one-ninth equals 3 to the power minus 2, we get x equals minus 2.
Now we turn to the three fundamental laws of logarithms. These laws are the tools that make logarithms so powerful for calculation.
The First Law, called the Product Law, states: the logarithm of a product at any non-zero base equals the sum of the logarithms of its factors at the same base.
Symbolically: logₐ(m × n) = logₐ m + logₐ n.
This extends to any number of factors: logₓ(m × n × p) = logₓ m + logₓ n + logₓ p.
A critical warning: the logarithm of a sum is NOT the sum of the logarithms. logₐ(m + n) is not equal to logₐ m + logₐ n. Similarly, the logarithm of a difference is not the difference of logarithms.
The Second Law, the Quotient Law, states: the logarithm of a fraction at any non-zero base equals the difference between the logarithm of the numerator and the logarithm of the denominator, both at the same base.
Symbolically: logₐ(m/n) = logₐ m − logₐ n.
Again, be careful: logₐ m / logₐ n is not equal to logₐ m − logₐ n. The first expression is actually the change of base formula, which we will meet later.
The Third Law, the Power Law, states: the logarithm of a power of a number at any non-zero base equals the logarithm of the number at the same base multiplied by the power.
Symbolically: logₐ(mⁿ) = n logₐ m.
An important corollary follows from this.
Since the nth root of m equals m to the power one over n, we have: logₐ ⁿ√m = (1/n) logₐ m. This shows how roots become simple divisions in the logarithmic world.
Logarithms to base 10 are so common that they have a special name: common logarithms. When no base is written, base 10 is always understood. So log 8 means log₁₀ 8, and log a means log₁₀ a.
From the power law, we can derive useful values. log₁₀ 100 = 2, since 100 is 10 squared. Similarly, log₁₀ 1000 = 3, and log₁₀ 10000 = 4. Also remember that log₁₀ 1 = 0 and log₁₀ 10 = 1.
Let us see how these laws allow us to expand complex expressions.
Suppose y equals (a⁴ × b²)/c³. Taking logarithms: log y = log(a⁴ × b²/c³). By the quotient law, this becomes log(a⁴ × b²) − log(c³). By the product law: log a⁴ + log b² − log c³. Finally, by the power law: 4 log a + 2 log b − 3 log c.
The process works in reverse too. Given log V = log π + 2 log r + log h − log 3, we can combine to get log V = log(πr²h/3), so V equals πr²h/3.
Here is a worked example using given values. Given that log 2 equals 0.3010 and log 3 equals 0.4771, let us find log 6. Since 6 equals 2 times 3, we have log 6 equals log 2 plus log 3, which equals 0.3010 plus 0.4771, giving 0.7781.
For log 5, we write 5 as 10 divided by 2. So log 5 equals log 10 minus log 2, which equals 1 minus 0.3010, giving 0.6990.
For log of root 24, we write this as half of log 24. Since 24 equals 2 cubed times 3, we get half of (3 × 0.3010 + 0.4771), which equals 0.69005.
Now we explore some deeper properties that reveal the elegant structure of logarithms.
First, the reciprocal property.
Since 2 cubed equals 8, we have log base 2 of 8 equals 3. But also, 8 to the power one-third equals 2, so log base 8 of 2 equals one-third. Notice that 3 and one-third are reciprocals. In general: log_b a = 1/log_a b. Equivalently: log_b a × log_a b = 1.
Second, the useful identity.
logₐ(aˣ) = x. Second, the useful identity: logₐ(aˣ) = x. This follows directly from the definition. For example, log base 2 of 2 to the fifth equals 5.
From this, we can derive that a^(logₐ m) = m. This is a powerful tool for simplifying expressions.
Third, the change of base formula.
We can express any logarithm in terms of another base: log_b a = (log_x a)/(log_x b), where a, b, and x are all positive.
For example, to find log base 100 of 1000, we write this as log₁₀ 1000 / log₁₀ 100, which equals 3 divided by 2, or 1.5.
Let us apply the change of base formula to evaluate log base 16 of 32 minus log base 25 of 125 plus log base 9 of 27. Converting each term: log 32 over log 16 becomes log of 2 to the fifth over log of 2 to the fourth, which equals 5 over 4. Similarly, log 125 over log 25 becomes log of 5 cubed over log of 5 squared, which equals 3 over 2. And log 27 over log 9 becomes log of 3 cubed over log of 3 squared, which equals 3 over 2. The expression becomes 5 fourths minus 3 halves plus 3 halves, which simplifies to 5 fourths or 1.25.
Here is a final elegant result to prove.
If 1 over log base a of x plus 1 over log base b of x equals 2 over log base c of x, then c squared equals a times b.
Using the reciprocal property, we rewrite: logₓ a + logₓ b = 2 logₓ c. By the product law, logₓ(ab) = logₓ(c²). Therefore, ab = c², as required.
Let us recap the key takeaways from this lesson.
First, logarithms and exponentials are inverse operations. aᵇ = c ⇔ logₐ c = b.
Second, the three fundamental laws. The product law turns multiplication into addition. The quotient law turns division into subtraction. The power law turns exponents into multiplication.
Third, common logarithms use base 10, and when no base is shown, 10 is understood.
Fourth, the reciprocal property. log_b a = 1/log_a b.
Fifth, the change of base formula allows us to compute any logarithm using a convenient base. log_b a = (log_x a)/(log_x b).
Sixth, logarithms transform complicated arithmetic operations into simpler ones. This was their original purpose in the age before calculators.
Logarithms are a beautiful example of how mathematics creates bridges between different forms of expression. By mastering these concepts and laws, you have added a powerful set of tools to your mathematical toolkit. Practice converting between forms, applying the laws, and simplifying expressions until these techniques feel natural. Thank you for your attention, and I look forward to seeing you apply these ideas with confidence.